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Len [333]
3 years ago
15

According to a study published in the New England Journal of Medicine, overweight people on low-carbohydrate and Mediterranean d

iets lost more weight and got greater cardiovascular benefits than people on a conventional low-fat diet (The Boston Globe, July 17, 2008). A nutritionist wishes to verify these results and follows 30 dieters on the low-carbohydrate and Mediterranean diets and 30 dieters on the low-fat diet. These data (measured in pounds) can be found on the text website, labeled Different Diets. Let Low-carb or Mediterranean and Low-fat diets represent populations 1 and 2, respectively. Use Table 2.
Low-carb Low-carb/

Mediterrean Diets Low- Fat diet Mediterreand diet Low-fat Diet

9.5 6.5 6.8 5.9
8.1 5.8 9.1 6.9
10.4 9.9 9.4 9.1
11.9 5.1 10.2 8.0
11.8 8.0 9.5 8.9
12.6 6.3 9.5 3.4
6.7 6.3 9.4 4.6
9.6 4.4 12.0 6.2
11.6 5.7 9.9 4.6
8.4 5.9 9.7 6.7
9.0 6.8 9.2 4.6
7.5 5.1 13.0 7.1
7.2 6.3 11.3 11.0
8.5 5.5 13.6 4.5
8.8 5.5 9.0 3.9


a.Set up the hypotheses to test the claim that the mean weight loss for those on low-carbohydrate or Mediterranean diets is greater than the mean weight loss for those on a conventional low-fat diet.

H0: ?1 ? ?2 ? 0; HA: ?1 ? ?2 > 0
H0: ?1 ? ?2 = 0; HA: ?1 ? ?2 ? 0
H0: ?1 ? ?2 ? 0; HA: ?1 ? ?2 < 0

b-1.Using the appropriate commands in Excel, find the value of the test statistic. Assume that the population variances are equal and that the test is conducted at the 5% significance level. (Round your answer to 2 decimal places.)

Test statistic


b-2. What is the critical value and the rejection rule?
Reject H0 if tdf < -1.672.
Reject H0 if tdf < -2.002.
Reject H0 if tdf > 2.002.
Reject H0 if tdf > 1.672.

c. At the 5% significance level, can the nutritionist conclude that overweight people on low-carbohydrate or Mediterranean diets lost more weight than people on a conventional low-fat diet?
No
Yes

Low-carb/Mediterrean Diets Low-fat Diet
9.5 6.5
8.1 5.8
10.4 9.9
11.9 5.1
11.8 8
12.6 6.3
6.7 6.3
9.6 4.4
11.6 5.7
8.4 5.9
9 6.8
7.5 5.1
7.2 6.3
8.5 5.5
8.8 5.5
6.8 5.9
9.1 6.9
9.4 9.1
10. 2 8
9.5 8.9
9.5 3.4
9.4 4.6
12 6.2
9.9 4.6
9.7 6.7
9.2 4.6
13 7.1
11.3 11
13.6 4.5
9 3.9
Mathematics
1 answer:
Fed [463]3 years ago
6 0

Answer:

Step-by-step explanation:

a)

Here researcher claims that the mean weight loss for those on  low-carbohydrate or Mediterranean diets is greater than the mean  weight loss for those on a conventional low-fat diet

Set the null and alternative hypotheses to test the above claim as below.

H_0:\mu_1\leq\mu_2\\\\=H_0:\mu_1-\mu_2\leq 0 versus H_A:\mu_1 > \mu_2\\\\H_A:\mu_1 -\mu_2>0

b)

Assume that the variance of the two populations are equal.

Under the null hypothesis, the test statistics is defined as

t= \frac{\bar x_1 - \bar x_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

where s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}

Use the foll owing Excel instructions and conduct the above test.

Step1: Enter the datainto two columns of spreadsheet.

Step2: Select Data Analysis from Data ribbon.

Step3: Select t-test. Two-sample Assuming equal variances.

Step4: Input the data range for variable 1 and variable 2.

Step5: Enter 0 as the Hypothesized Mean Difference.

Step6: Click OK.

Thus, the resultant outout is as follows:

         

                                                       Low-carb                        Low-fat

Mean                                               9.773333333       6.283333333

Variance                                          3.197885057                    3.160747126

Observations                                  30                                      30

Pooled Variance                             3.179316092

Hypothesized Mean Difference     0

df                                                      58

t Stat                                                 7.580610844

P(T < = t) one-tail                              1.54916E - 10

t Critical one — tail                           1.671552762

P(T <2) two-tail                                  3.09831E - 10

t Critical two - tail                              2.001717484

1) From the above output, the test statistics is obtained as t =  7.5806

2) From the above output, the critical value for one-tailed testis  obtained as t_{crit}=1.672

Rejection rule:  Reject H_0 if t_{crit}>1.672

c)

At 5% level of significance, the calculated value of test statistics  is greater than the critical value.

Therefore, reject the null hypothesis.

Hence, the nutritionist conclude that overweight people on  low-carbohydrate or Mediterranean diets lost more weight than  people on a conventional low-fat diet.

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Answer:

No, George does not have enough money to buy everything.

The correct expression is:

15 - 9 - 2.65 - 1.35 - 2(1.74)

Step-by-step explanation:

Given that:

Money present with George initially = $15

Money required to see the movie = $9

Money required to buy a pretzel = $2.65

Money required to buy a drink = $1.35

Money required to buy 2 veggie cups = $2(1.74)

To find:

Whether there is enough money with George to complete all the tasks.

Solution:

First of all, let us calculate all the money required and then we can subtract all the money from the initial amount to find whether there is enough money present with George.

If the result is in positive, there is enough money with George.

Total money required for all the activities = $9 + $2.65 + $1.35 + $2(1.74) = $16.48

Expression = $15 - $16.48 = -$1.48

The result is negative, therefore there is not enough money.

The correct expression is:

15 - 9 - 2.65 - 1.35 - 2(1.74)

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<h3>Slope Intercept Form</h3>

We are given the relationship between the number of costumes that Kai made, x, and the total length of fabric that he purchased y, in yards as; y - 5x = 2

Where;

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Thus, our equation can be rewritten as;

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